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### Course: Integral Calculus>Unit 1

Lesson 17: Integration by parts

# Integration by parts review

Review your integration by parts skills.

## What is integration by parts?

Integration by parts is a method to find integrals of products:
$\int \phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}u\left(x\right){v}^{\prime }\left(x\right)dx=u\left(x\right)v\left(x\right)-\int \phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}{u}^{\prime }\left(x\right)v\left(x\right)dx$
or more compactly:
We can use this method, which can be considered as the "reverse product rule," by considering one of the two factors as the derivative of another function.

## Practice set 1: Integration by parts of indefinite integrals

Let's find, for example, the indefinite integral $\int x\mathrm{cos}x\phantom{\rule{0.167em}{0ex}}dx$. To do that, we let $u=x$ and $dv=\mathrm{cos}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$:
$\int x\mathrm{cos}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=\int u\phantom{\rule{0.167em}{0ex}}dv$
$u=x$ means that $du=dx$.
$dv=\mathrm{cos}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$ means that $v=\mathrm{sin}\left(x\right)$.
Now we integrate by parts!
$\begin{array}{rl}\int x\mathrm{cos}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx& =\int u\phantom{\rule{0.167em}{0ex}}dv\\ \\ & =uv-\int v\phantom{\rule{0.167em}{0ex}}du\\ \\ & =x\mathrm{sin}\left(x\right)-\int \mathrm{sin}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx\\ \\ & =x\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)+C\end{array}$
Problem 1.1
$\int x{e}^{5x}dx=?$

Want to try more problems like this? Check out this exercise.

## Practice set 2: Integration by parts of definite integrals

Let's find, for example, the definite integral ${\int }_{0}^{5}x{e}^{-x}dx$. To do that, we let $u=x$ and $dv={e}^{-x}\phantom{\rule{0.167em}{0ex}}dx$:
$u=x$ means that $du=dx$.
$dv={e}^{-x}\phantom{\rule{0.167em}{0ex}}dx$ means that $v=-{e}^{-x}$.
Now we integrate by parts:
$\begin{array}{rl}& \phantom{=}{\int }_{0}^{5}x{e}^{-x}\phantom{\rule{0.167em}{0ex}}dx\\ \\ & ={\int }_{0}^{5}u\phantom{\rule{0.167em}{0ex}}dv\\ \\ & =\left[uv{\right]}_{0}^{5}-{\int }_{0}^{5}v\phantom{\rule{0.167em}{0ex}}du\\ \\ & =\left[-x{e}^{-x}{\right]}_{0}^{5}-{\int }_{0}^{5}-{e}^{-x}\phantom{\rule{0.167em}{0ex}}dx\\ \\ & =\left[-x{e}^{-x}-{e}^{-x}{\right]}_{0}^{5}\\ \\ & =\left[-{e}^{-x}\left(x+1\right){\right]}_{0}^{5}\\ \\ & =-{e}^{-5}\left(6\right)+{e}^{0}\left(1\right)\\ \\ & =-6{e}^{-5}+1\end{array}$
Problem 2.1

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• Thanks Sal. I am the master of integration by parts. Do not test me. I am now the best, I think I should do the testing.
• in the int (0 -> pi) of xsin(2x)dx problem, in the solution, the third to last line, shouldn't that be (sin(2x)/4) not (sin(4x)/4)? or am I missing something?
• You are correct, it is a typo, though it does not effect the result since sin(nπ) = 0 for all integers n.
You can, and should, report the error since you found it.
• Why hasn't Sal explained about the compact form of Integration by parts??i don't understand it!! It contradicts to what Sal said about differentials earlier that the differentials are not numbers or function which can't cancelled or algebraically manipulated!!
• The "compact form" is just a different way to write the form used in the videos. Basically, the only difference is that the "video form" uses prime notation (f'(x)), and the "compact form" uses Leibniz notation (dy/dx). If you are used to the prime notation form for integration by parts, a good way to learn Leibniz form is to set up the problem in the prime form, then do the substitutions f(x) = u, g'(x)dx = dv, f'(x) = v, g(x)dx = du. At least, that's how it clicked for me.

As far as the manipulating differentials goes, it's true that you can't just treat differentials like they are normal terms in an equation (as if dx were the variable d times the variable x), but it is legal to split up the dy/dx when differentiating both sides of an equation. The concept here is exactly the same as what is used when doing u-substitution (URL to video below if you need it).
Hope this helps, and good luck with your work!
• Like Bhoovesh I am also fuzzy about the compact notation. It seems that the confusion is not with Leibniz notation vs Newton's, but rather I am concerned about falling in a pit as a consequence of having only one letter in an expression for which I am accustomed to two. The dropping of the x's and dx's makes me nervous about getting fouled up as a consequence of x not being the only variable. Where is y, and how does one keep track of it with the more compact notation? I would like to know the conventions and rules for this.
• Hi. There is a gap of explanation about the compact notation. I didn't catch why 'dx' becomes 'dv' and 'du'. Thank you.
• In case anyone else is confused,
to find 'dv' from 'v', take the derivative and multiply by 'dx'. You can think of this as because the derivative of 'v' is 'dv/dx', so given dv/dx = a => dv = adx. This is similar to one way of doing u-substitution.
• Why does the integral of e^5x dx = 1/5 e^5x? Is it an application of the reverse chain rule?
Thanks very much!
• That's one way of thinking about it, yes. As you continue on in math, this will become almost second-nature and you won't even think about the chain rule when integrating simple exponential functions.
• What is the use of integration? When do we use it in our practical lives?
• we can use integration to find Displacement from Velocity, and Velocity from Acceleration, Voltage across a Capacitor, Moments of Inertia by Integration and many more...
• How would you integrate something in parentheses, like (x^2 +1)^1/2?
• That depends hugely on what's in the parentheses. √(x²+1) can be done by trigonometric substitution, but √(x³+1) cannot be done by elementary means.
• ILATE technique will be useful
• how do we chose what u equals and what dV equals?
• Usually, it depends on what’s easier to integrate, or what simplifies the problem more.

In general, you can remember the acronym LIATE - Log, Inverse trig, Algebraic (power functions), Trig, Exponential. When using integration by parts, you’ll want to choose u to be whatever comes first in that acronym (and the other function will become dv). This helps you pick which function is easier to integrate.

Example: if you had the integral of xln(x) dx, you would choose ln(x) to be u and x dx to be dv. See how this avoids having to find the anti-derivative of ln(x)

Hope this helps!